# equation for the rate determining step

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#1
hello

rate equation = k[NO]^2 [H2]

overall reaction = 2NO + 2H_2 -> N_2 + 2H_2O

The first step of a three-step reaction mechanism is shown.
2NO -> N_2O_2
The second step of the mechanism is the rate determining step.
Suggest an equation for the rate determining step.

what is this question.......

the answer says it must have N_2O_2 and H_2 on LHS
since The rate equation must include all the species up to and including the rate determining step

but the problem is N_2O_2 is not involved in rate equation

the compound in the rate equation is [NO]^2 !!!!
what?
I just totally lost
0
2 months ago
#2
(Original post by universalcj)
hello

rate equation = k[NO]^2 [H2]

overall reaction = 2NO + 2H_2 -> N_2 + 2H_2O

The first step of a three-step reaction mechanism is shown.
2NO -> N_2O_2
The second step of the mechanism is the rate determining step.
Suggest an equation for the rate determining step.

what is this question.......

the answer says it must have N_2O_2 and H_2 on LHS
since The rate equation must include all the species up to and including the rate determining step

but the problem is N_2O_2 is not involved in rate equation

the compound in the rate equation is [NO]^2 !!!!
what?
I just totally lost
Which spec are you doing?
0
#3
(Original post by Pigster)
Which spec are you doing?
the new one
0
2 months ago
#4
(Original post by universalcj)
the new one
AQA? OCR A/B? Pearson? CIE? etc.
0
#5
(Original post by Pigster)
AQA? OCR A/B? Pearson? CIE? etc.
Pearson IAL
0
1 month ago
#6
(Original post by universalcj)
Pearson IAL
N2O2 is allowed in the RDS.

N2O2 isn't in the overall equation. It is generated by the 1st (fast) step and destroyed in the 2nd (RDS) step.

If you added N2O2, then it would be 1st order WRT [N2O2].

You don't add N2O2, it is made in the first step. If you double [NO] in the 1st step, you will quadruple the rate at which N2O2 is made, effectively quadrupling [N2O2], which (since it is 1st order) will quadruple the rate of the RDS, i.e. doubling [NO] leads to a quadrupling of the rate, which rather sounds like it is 2nd order WRT [NO].

If you don't like the sound of that, consider A + B -> where rate = k [A] [B]. Doubling [A] leads to a doubling of rate, ditto [B]. Doubling both leads to 2x then 2x, i.e. 4x rate. Why is that any different to A + A -> B, if you 2x [A] (the left one) then you automatically 2x [A] (the right one) and hence 2x [A] must 4x rate.
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